Can you find the treasure??

Question

My big bro gave this problem one week ago. I could not still solve it.Please HELP.

STORY

A man was just looking for items in his store room. Suddenly he found a map , which showed

then it stated That if the man goes straight from the pole(P) to house A and turns 90* and moves to M such that PA=AM. Similarly if he goes from P to B and turns 90* again to move from B to N such that BN=PN

THEN a straight line MN is produced.In the midpoint of MN the TREASURE IS PRESENT.

THE PROBLEM

Now the man went to find the treasure but when he reached there he was shocked. because the pole(P) was cut down (i.e the pole was absent from the place)

NOW CAN HE FIND THE TREASURE????

thanks in advance!!

Answer 1

As the diagram suggests, any starting point and the endpoints ($A^\prime$ and $B^\prime)$ of the routes through $A$ and $B$ determine pairs of congruent triangles with the (other) vertices $X$ and $Y$ of the square with diagonal $\overline{AB}$. The key midpoint property then becomes clear (and nicely related to the distance from starting point to square-corner).

Now, starting at any point on the $Y$ side of $\overleftrightarrow{AB}$, the instructions (with appropriately-oriented turns) take you to $X$ (necessarily the midpoint of $\overline{A^\prime B^\prime}$); and vice-versa. (What about points on $\overleftrightarrow{AB}$?) So, you should check both points, just to be sure. $\square$

Answer 2

It doesn't matter. He can put down a new pole wherever he wants and follow the same instructions. He will still find the treasure.

Answer 3

Identify the euclidean plane $\mathbb{R}^2$ with the complex plane $\mathbb{C}$.

Let $T$ be the position of the treasure.

If the direction of rotation is the one given in the picture, then

$$\begin{cases} M &= A - i (A-P)\\ N &= B + i (B-P) \end{cases} \implies T = \frac12(M+N) = \frac{A+B}{2} - i\frac{A-B}{2}$$

This means in order to get to $T$, one first move to the midpoint of $AB$, rotate $90^\circ$ clockwise and then move half the distance between $AB$. Alternatively, if one construct a square with $AB$ as its diagonal, the treasure $T$ will be located on one of the corners of the square.

Answer 4

The answer is that he will have to dig in two places. The places are found by bisecting the line $AB$, and on the perpendicular bisector of $AB$ marking of the distances $AB/2$ from the midpoint of $AB$ in either direction.

The proof:

By rescaling and rotating, we can place $A$ and $B$ at $(-1,0)$ and $(1,0)$ in a cartesian coordinate system. Now the pole was at some other location; say in the lower half plane at angle $OAP = \theta$ and length $AP = \alpha$. Then $$P = (-1+\alpha \cos \theta, -\alpha \sin \theta)$$ And by drawing a line of lenght $\alpha$ from $A$ perpendicular to $AP$ you find $M$ at $$M=(1+\alpha \sin \theta, \alpha \cos \theta)$$ (Actually, these formulas also work if $P$ is in the upper half plane; then $\theta$ is negative.)

Now if lenght $PB = \beta$, you can (but don't need to) determine $\beta$ as $\sqrt{5-4 \cos \theta}$. And if angle $ABP$ is $\phi$, then $$\sin \phi = \frac{\alpha \sin \theta}{ \beta}$$ and $$\cos\phi = \frac{2-\alpha \cos \theta}{ \beta}$$ Next move to $N$ at $$N = (1,0) + (-\beta \sin \phi, \beta \cos \phi) = (1-\alpha \sin \theta, 2-\alpha \cos \theta)$$ Finally the treasure $T$ is at $$T = \frac{M+N}{2} = (0,1)$$

So $T$ is at a corner of the square whose diagonal is $AB$; but there are two such corners so you have to dig in two places (unless you get lucky on the first hole).